Convexity, linearity and their combination for MLE I'm going through Murphy's ML a Probabilistic Perspective book and in chapter 9 we have the following excerpt talking about the MLE of exponential family distributions:

My question is: How do we arrive at the conclusion that because -A(θ) is concave in θ and $\mathbf{\theta}^T\phi(D)$ is linear in θ, their difference will also be concave?
What's the relationship between concavitiy and linearity here?
 A: $-A$ is concave if
$$
-\alpha A(\boldsymbol\theta_1) - (1-\alpha)A(\boldsymbol\theta_2) \leq -A(\alpha\boldsymbol\theta_1 + (1-\alpha)\boldsymbol\theta_2)
$$
for $0 \leq \alpha \leq 1$, and $\boldsymbol\theta_1, \boldsymbol\theta_2$ in the domain of $A$.
On the other hand
$$
\alpha(\boldsymbol\theta_1^T\phi(\mathcal{D}) - A(\boldsymbol\theta_1)) + (1-\alpha)(\boldsymbol\theta_2^T\phi(\mathcal{D}) - A(\boldsymbol\theta_2)) 
$$
$$
=\alpha\boldsymbol\theta_1^T\phi(\mathcal{D}) - \alpha A(\boldsymbol\theta_1) + (1-\alpha)\boldsymbol\theta_2^T\phi(\mathcal{D}) -(1-\alpha) A(\boldsymbol\theta_2)
$$
$$
=\alpha\boldsymbol\theta_1^T\phi(\mathcal{D}) + (1-\alpha)\boldsymbol\theta_2^T\phi(\mathcal{D}) - \alpha A(\boldsymbol\theta_1) -(1-\alpha) A(\boldsymbol\theta_2)
$$
$$
\leq (\alpha\boldsymbol\theta_1 + (1-\alpha)\boldsymbol\theta_2) ^T\phi(\mathcal{D}) - A(\alpha\boldsymbol\theta_1 + (1-\alpha)\boldsymbol\theta_2)
$$
The last inequality is easily verified, as the first addend of the last line is linear and the second one corresponds to the inequality we proposed initially.
A: The relationship is that a linear function is both convex and concave, so we have that $\theta^T \phi(D)$ is concave, and $-NA(\theta)$ is concave. Thus the loglikelihood is a sum of two concave functions, which is a concave function (that one is an easy proof). 
For completeness, let us write out the proof that a sum of two concave functions is concave.  Let $f,g$ be concave on their common domain $D$.  Then, by definition of concavity, we have that $f(\alpha x +(1-\alpha) y) \ge \alpha f(x) + (1-\alpha) f(y)$, for all $0 \le \alpha \le 1$ and $x,y \in D$, likewise for $g$.  Then we can compute 
\begin{align} 
(f+g)(\alpha x + (1-\alpha) y) &= f(\alpha x + (1-\alpha) y) +g(\alpha x + (1-\alpha) y) \\
     &\le \alpha f(x) + (1-\alpha) f(y) + \alpha g(x) + (1-\alpha) g(y) \\
   &=  \alpha (f+g)(x) + (1-\alpha) (f+g) (y)
\end{align}
showing indeed that the susm $f+g$ is concave.
